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In the book there is this exercise:

Let E be an extension fiel of F, with $\alpha, \beta \in E$. Suppose $\alpha$ is transcendental over F but algebraic over $F(\beta)$. Show that $\beta$ is algebraic over $F(\alpha)$.

The suggested solution was: enter image description here

It doesn't seem that they use that $\alpha$ is transcendental in F, or do they? Or is the statement valid without this assumption?

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  • $\begingroup$ There is a polynomial P($\alpha$) = 0 with coefficients polynomials f($\beta$). You do just rewrite this polynomial to conclude. $\endgroup$ – Piquito Apr 23 '15 at 18:17
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Actually, they use that hypothesis in the part

a polynomial in $\alpha$ with coefficients that are polynomials in $\beta$ can be formally rewritten as a polynomial in $\beta$ with coefficients that are polynomials in $\alpha$

to conclude that this new polynomial (seen in $(F[\alpha])[x]$) is not the zero polynomial. This is necessary to conclude that $\beta$ satisfy a non-zero polynomial over $F(\alpha)$ so $\beta$ is algebraic over $F(\alpha)$.

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  • $\begingroup$ @AlexWertheim You're completely right! I'm sorry, I'll erase it right now. I think a counterexample goes that way. $\endgroup$ – Daniel Apr 23 '15 at 17:59
  • $\begingroup$ Of course Alex. It was certainly a lapsus of Daniel. $\endgroup$ – Piquito Apr 23 '15 at 18:06
  • $\begingroup$ @LuisGomezSanchez That's right. On the other hand, can you think a counterexmaple? I'm thinking and I can't find anything. $\endgroup$ – Daniel Apr 23 '15 at 18:09
  • $\begingroup$ @DanielEscudero: doing so might be difficult. You'd have to ensure the minimal polynomial of $\alpha$ over $F$ divides all the coefficients of the polynomial obtained by rewriting as polynomial in $\beta$. I can't see of any easy way to rig this. Roots of unity might be a good place to start, but that's just an offhand guess. $\endgroup$ – Alex Wertheim Apr 23 '15 at 18:12
  • $\begingroup$ Is the point that when we reqrite the poluynomial we have to make sure that not coefficients infront of each $\beta$ is zero? So we are using that $\alpha$ is transcendental in F, on each coefficient, not on the polynomial as a whole? $\endgroup$ – user119615 Apr 23 '15 at 18:25

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